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Let $A\in\mathbb{R}^{m\times n}$. Is $A$ invertible if and only if $A^\intercal A$ is invertible?

First of all, this statement makes sense only if $m=n$ since only square matrices are invertible.

So let's assume $m=n$. In this case, if $A$ is invertible, its columns are linearly independent. So this means that the rows of $A^\intercal$ are linearly independent, meaning that $A^\intercal$ is invertible.

Thus $(A^\intercal A)^{-1}=A^{-1}(A^\intercal)^{-1}$ exists.

And vice versa, if $A^\intercal A$ is invertible, this means that $A^{-1}$ needs to exist.

Correct?

And for $m\neq n$ the statement is false, right?

Rhjg
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2 Answers2

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For any matrix $A$ it holds that $$\operatorname{rank}(A^T A) = \operatorname{rank}(A). $$ For $m=n$ this implies that $A^T A$ is invertible iff $A$ is.

LordOfNumbers
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Your proof of $A$ being invertible implying $A^\top A$ being invertible is valid. On the other hand, if $A$ is not invertible, then $\det A = 0$. It follows directly that $\det (A^\top A) = \det(A) \det(A^\top) = \det(A)^2 = 0$, showing $A^\top A$ is not invertible.

Ethan
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  • This also means that if the columns of A are linearly dependent, then $A^\intercal A$ is not invertible. – Rhjg Mar 21 '22 at 20:57
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    $\det A$ only makes sense if $m=n$. – Ivo Terek Mar 21 '22 at 21:04
  • Thats my point exactly. For $m\neq n$, the statement $A$ invertible if and only if $A^\intercal A$ invertible does not make sense. – Rhjg Mar 21 '22 at 21:06